Good Morning. I haven't done probability in a while and a friend's son asked me how to a problem:
Let $Y$ denote the number of broken eggs in a randomly selected carton of one dozen "store brand" eggs at a local super market. Suppose that the probability distribution of $Y$ is a follows"
$$\begin{matrix}
y:&0&1&2&3&4\\
p:&0.78&0.11&0.07&0.03&0.01\\
\end{matrix}$$
A quality control inspector at the store keeps looking at randomly selected cartons of eggs until he finds one with at least 2 broken eggs. Find the probability that this happens in one of the first three cartons he inspects.
Again, I haven't done probability in a while. My first thoughts is that the probability that i should use is $p=.11$, since we are talking at least $2, (2+3+4)$. My second thought is to do a binomial with $n=3, k=1$, since the problem is saying $1$ out of the first $3$ cartons he inspects. Thus
$$P(Z=1)=\binom{3}{1}.11\cdot .89^2\approx .2614$$
However, the problem clearly says that the inspector works until he finds the carton with the necessary amount of broken eggs; in this case, 2 or more. This makes me think it's geometric since it's time to failure. So
$$P(Z=3)=(.89)^3\cdot.11\approx .0871$$
Both of these answers are incorrect as the correct answer is given as $.295$. Can anyone tell me what am doing wrong here?
Best Answer
If it happens in the first three tries, then either he finds one on the first, or he fails and finds one on the second or he fails fails and finally finds one on the third. Thus \begin{align*} P(Z \leq 3) &= P(Z = 1)+P(Z=2)+P(Z = 3)\\ &=.11 +(1-.11)(.11)+(1-.11)^2(.11) \\ &=0.295031, \end{align*} where you can split the probability since these are disjoint events. And yes, $$Z\sim \text{Geom}(p = .07+.03+.01 = .11).$$